$A$ be a $n \times n$ matrix with $\text{rank}\,(A)\lt n$: multiple choice question 
Let $A$ be an $n \times n$ real non-zero matrix of rank less than $n$. Then one of the following is true? :
(A) there exists an $n \times n$ real non-zero matrix $B$ such that $BA = 0$.
(B) there may not always exist an $n \times n$ real non-zero matrix $B$ such that $BA = 0$.
(C) there exists an $n \times n$ real non-zero matrix $B$ such that $BA = I$.
(D) if $B$ is such that $BA = 0$, then $AB = 0$.

My Attempt: Consider a $2\times 2$ matrix $A$ of the form  $\begin{pmatrix}
0 &1 \\ 
0 & 0
\end{pmatrix}$ 
and a $2\times 2$ matrix $B$ of the form  $\begin{pmatrix}
0 &a \\ 
0 & c
\end{pmatrix},$ where $a,c$ are any non zero real numbers. 
Then I have that $BA=0$ holds. So option $(A)$ may hold. 
Option $(B)$ can also hold if we take $2\times 2$ matrix $B$ of the form  $\begin{pmatrix}
1 & a \\ 
2 & c
\end{pmatrix}$
$(C)$ is clearly false. 
Also $(D)$ is also false as is evident if we take $2 \times 2$ matrix $A$ of the form  $\begin{pmatrix}
0 &1 \\ 
0 & 0
\end{pmatrix}$ and $2 \times 2$ matrix $B$ of the form  $\begin{pmatrix}
0 &a \\ 
0 & c
\end{pmatrix}$
So I am confused between $(A)$ and $(B)$. Which one should be the right choice?
 A: Showing existence of a matrix $B$ such that $AB = 0$ requires only one such matrix (whatever the value of $n$.)
You shown such existence (option (A)) for $n = 2$. 
Hint: Can you generalize to consider other $n\times n$ matrices A whose rank is less than n? If yes, that is, if you can show existence of "even one" matrix $B$ such that $BA = 0$ whatever the value of $n$, then (A) is the option to choose.
Note that since the rank of $A$ is less than $n$, (by hypothesis), the columnspace given by $\text{im} A=\{Ax:x\in\Bbb R^n\}$ is therefore not of full dimension. 
So choose any matrix $B$ which maps $\text{im} A$ to zero but sends a vector out from $\text{im} A$ to some non-zero vector. 
Such a $B$ can certainly be constructed for any $n\times n$ matrix $A$ with rank less than $n$. 
So we have generalized to show existence of a $B$ such that $BA = 0$ whatever the dimension $n$.
A: If rank of $A$ is less than $n$, then the columnspace ($im A=\{Ax:x\in\Bbb R^n\}$) is not full dimension, so any $B$ which maps $im A$ to zero but sends a vector $v\notin im A$ to a nonzero vector, will do the job.
